Name: Collision Between a Point Mass and a Rigid Body
Uploaded: 2019-07-19
Duration: 7 min 57 s
Description: Can we treat the collision of a point mass with a rigid body in the same way we treated the collision of point masses? Watch this video to know about it.

Question

Each of the double pulleys shown has a centroidal mass moment of inertia of ${mr}^{2},$ inner radius r and an outer radius $2 r .$ Assuming that the bearing friction of hinge at A and at B is equivalent to torque of magnitude $\frac{mgr}{4}$ then the tension (in N) in the string connecting the pulleys is : $(m = 3 kg, g = 10 m s^{- 2}, and r = 0.1 m)$

Accepted Answer

Given,
Mass of block $(m) = 3 kg$
Inner radius $(r) = 0.1 m$

Suppose acceleration of block $C$ is $a$ and angular acceleration of pulley $A$ and $B$ is $α_{1}$ and $α_{2}$ respectively.

From $F . B . D .$ of Block $C -$
$m g - T_{1} = m a 30 - T_{1} = 3 a \dots \dots (1)$

Net torque about centroidal axis
$(T_{1} \times r) - (T_{2} \times 2 r) - (τ_{f_{r}}) = I_{1} α_{1} (T_{1} - 2 T_{2}) r - \frac{m g r}{4} = m r^{2} α_{1} T_{1} - 2 T_{2} - \frac{m g}{4} = m r α_{1} T_{1} - 2 T_{2} - \frac{30}{4} = 0.3 α_{1} \dots \dots (2)$

Net torque about centroidal axis
$T_{2} \times r - (τ_{f_{r}}) = I_{2} α_{2} T_{2} r - \frac{m g r}{4} = m r^{2} α_{2} T_{2} - \frac{m g}{4} = m r α_{2} T_{2} - \frac{30}{4} = 0.3 α_{2} \dots \dots (3)$

From no slip condition,
$a = r α_{1} a = 0.1 α_{1} \dots \dots (4)$

When string connecting both the pulleys will move, both pulley's contact point will move by same linear distance. Using this constraint relationship-
$r \times θ_{1} = 2 r \times θ_{2}$

Double differentiating above relations-
$r α_{1} = 2 r α_{2} α_{1} = 2 α_{2} \dots \dots (5)$

Solving equations $(1), (2), (3), (4) and (5) -$
Tension in string connecting both the pulleys $(T_{2}) = 10 N$

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